# Solid formed after revolving the area between #y = x^2# and #y = x+2# around #y# axis?

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Two functions

#y=x^2# and #y=x+2#

Around y axis

It will be #(16π)/3# because there is a solid inside other or we have to considerer solid 1 plus solid 2

#(5π)/6 + (16π)/3 = (37π)/6#

Two functions

Around y axis

It will be

##### 1 Answer

Aug 6, 2018

graph{(y-x^2)(y-x-2)=0 [-7.37, 10.41, -1.955, 6.934]}

Ignore the

That leaves:

With